Question

sin 10° + sin 20° + sin 40° + sin 50º = sin 70° + sin 80°.​

Answer 1

Step-by-step explanation:

L.H.S. = 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50]

= 2cos5 (sin15+sin45)

= 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

= 2cos5 (2 x 1/2 x cos15)

= 2cos5 cos15

R.H.S. = sin70+sin80

= 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

sin75 = sin(90-15) = cos 15

L.H.S = 2cos5 cos15

R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ]

Answer 2

$\huge\mathtt\blue{Answer:-}$

Given as sin 50° + sin 10° = cos 20°

Let us consider the LHS sin 50° + sin 10°

On using the formula,

sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

sin 50° + sin 10° = 2 sin (50° + 10°)/2 cos (50° – 10°)

= 2 sin 60°/2 cos 40°/2

= 2 sin 30° cos 20

° = 2 × 1/2 × cos 20°

= cos 20°

= RHS

Thus proved